Problem: Rewrite the equation by completing the square. $4 x^{2} +28 x +49 = 0$ $(x + $
Answer: $\begin{aligned} 4 x^2 +28 x +49&=0 \\\\ 4 x^2 +28 x &=-49 \\\\ x^2 +7 x&=-\dfrac{49}{4} \end{aligned}$ Now we want to complete $x^2 +7 x$ into a perfect square. To do that, we should add $\left(\dfrac{{7}}{2}\right)^2={\dfrac{49}{4}}$ to it: $x^2{+7}x + {\dfrac{49}{4}}=\left(x +\dfrac{7}{2} \right)^2$ $\begin{aligned} x^2 +7 x&=-\dfrac{49}{4} \\\\ x^2 +7 x + {\dfrac{49}{4}}&=-\dfrac{49}{4} + {\dfrac{49}{4}} \\\\ \left(x +\dfrac{7}{2} \right)^2&=0 \end{aligned}$ In conclusion, the equation after completing the square is written as: $\left(x +\dfrac{7}{2} \right)^2=0$